早教吧作业答案频道 -->其他-->
已知H2(g)+12O2(g)=H2O(l);△H=─285.8KJ/moL可通过两种途径来完成分步进行的热效应如下:(1)H2(g)=2H△H1=+431.8KJ/moL(2)12O2(g)=O(g)△H2=+244.3KJ/moL(3)2H(g)+O(g)=H2O(g)△H
题目详情
已知H2(g)+
O2(g)=H2O (l);△H=─285.8KJ/moL可通过两种途径来完成分步进行的热效应如下:

(1)H2(g)=2H△H1=+431.8KJ/moL
(2)
O2(g)=O (g)△H2=+244.3KJ/moL
(3)2H (g)+O (g)=H2O (g)△H3=______
(4)H2O (g)→H2O (l)△H=-44kJ/moL.
| 1 |
| 2 |

(1)H2(g)=2H△H1=+431.8KJ/moL
(2)
| 1 |
| 2 |
(3)2H (g)+O (g)=H2O (g)△H3=______
(4)H2O (g)→H2O (l)△H=-44kJ/moL.
▼优质解答
答案和解析
H2(g)+
O2(g)=H2O (l)△H=-285.8KJ/moL,依据盖斯定律,反应焓变和起始物质和终了物质有关于变化过程无关,所以根据盖斯定律
(1)H2(g)=2H△H1=+431.8KJ/moL
(2)
O2(g)=O (g)△H2=+244.3KJ/moL
(3)2H (g)+O (g)=H2O (g)△H3
(4)H2O (g)→H2O (l)△H=-44kJ/moL
(1)+(2)+(3)+(4)=△H=-285.8KJ/moL
则(+431.8KJ/moL)+(+244.3KJ/moL)+△H3+(-44kJ/moL)=-285.8KJ/moL,
解得△H3=-917.9KJ/mol;
故答案为:-917.9KJ/mol.
| 1 |
| 2 |
(1)H2(g)=2H△H1=+431.8KJ/moL
(2)
| 1 |
| 2 |
(3)2H (g)+O (g)=H2O (g)△H3
(4)H2O (g)→H2O (l)△H=-44kJ/moL
(1)+(2)+(3)+(4)=△H=-285.8KJ/moL
则(+431.8KJ/moL)+(+244.3KJ/moL)+△H3+(-44kJ/moL)=-285.8KJ/moL,
解得△H3=-917.9KJ/mol;
故答案为:-917.9KJ/mol.
看了 已知H2(g)+12O2(g...的网友还看了以下: